Reversing the order of integration

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The discussion focuses on reversing the order of integration for a double integral involving the function f(x). The original integral is defined as ∫∫ f(x) dx dy, with bounds from x=y²/a to x=y and y=0 to y=a. The proposed new order is ∫∫ f(x) dy dx, with inner limits from y=sqrt(xa) to y=x and outer limits from x=0 to x=a. The community member confirms that the proposed solution is mostly correct but notes that the inner limits are reversed.

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mikeyrichster
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Hi I am trying to evaluate a line integral, and i need to reverse the order of integration, i'll call the function f(x) as it doesn't matter too much. and the bounds are:

double integral f(x) dxdy where inner integral is from x=y^2/a to x=y and the outer integral is from y=0 to y=a (where a is a positive constant)

talk about a headache! I've done the best i can and i came up with:

double integral fx dydx where inner integral is from y=sqrt(xa) to y=x and the outer integral is from x=0 to x=a

but i don't think this is right because when i use this for f(x) i get an impossible answer (ill spare you the details)

Any help would be appreciated!
 
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Hi mikeyrichster! Welcome to PF! :smile:

(have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
mikeyrichster said:
Hi I am trying to evaluate a line integral …

erm :redface:

… noooo! :biggrin:
double integral f(x) dxdy where inner integral is from x=y^2/a to x=y and the outer integral is from y=0 to y=a (where a is a positive constant)

talk about a headache! I've done the best i can and i came up with:

double integral fx dydx where inner integral is from y=sqrt(xa) to y=x and the outer integral is from x=0 to x=a
The area is from (0,0) to (a,a), to the right of the curve x = y2/a, and to the left of the the straight line,

which is the same as above the straight line and below the curve,

so your solution looks ok to me (except that the inner limits are the wrong way round). :smile:
 

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